The Spectrum of Periodic Schrödinger Operators
§I The Main Idea
Let Γ be a lattice of static ions in IRd . Suppose that the ions generate an electric
potential V (x) that is periodic with respect to Γ. Then the Hamiltonian for a single electron
moving in this lattice is
2
1
H = 2m
i∇
∇ + V (x)
This Hamiltonian commutes with all of the translation operators
(Tγ φ (x) = φ(x + γ )
γ∈Γ
Problem S.1 Prove that
i) Tγ is a unitary operator on L2 (IRd ) for all γ ∈ IRd .
ii) Tγ Tγ ′ = Tγ +γγ ′ for all γ , γ ′ ∈ IRd
Problem S.2 Prove that
∂T ϕ
∂ϕ
i) Tγ ∂x
= ∂xγi for all differentiable functions ϕ on IRd , 1 ≤ i ≤ d and γ ∈ IRd
i
ii) Tγ V ϕ = V Tγ ϕ for all γ ∈ Γ, all functions V that are periodic with respect to Γ and
all functions ϕ on IRd .
Pretend, for the rest of §I, that H and the Tγ ’s are matrices. We’ll give a rigorous
version of this argument later. We know that for each family of commuting normal matrices,
like {H, Tγ , γ ∈ Γ}, there is an orthonormal basis of simultaneous eigenvectors. These
eigenvectors obey
Hφα = eα φα
Tγ φα = λα,γγ φα
∀γγ ∈ Γ
for some numbers eα and λα,γγ .
As Tγ is unitary, all its eigenvalues must be complex numbers of modulus one. So
there must exist real numbers βα,γγ such that λα,γγ = eiβα,γγ . By Problem S.1.ii,
Tγ Tγ ′ ϕα = Tγ +γγ ′ ϕα
= eiβα,γγ +γγ ′ ϕα
= Tγ eiβα,γγ ′ ϕα = eiβα,γγ eiβα,γγ ′ ϕα = ei(βα,γγ +βα,γγ ′ ) ϕα
c Joel Feldman.
2000. All rights reserved.
1
which forces
∀γγ , γ ′ ∈ Γ
βα,γγ + βα,γγ ′ = βα,γγ +γγ ′ mod 2π
Thus, for each α, all βα,γγ , γ ∈ Γ are determined, mod 2π, by βα,γγ i , 1 ≤ i ≤ d. Given any d
numbers β1 , · · · , βd the system of linear equations (with unknowns k1 , · · · , kd )
that is
d
P
γ i · k = βi
1≤i≤d
γi,j kj = βi
1≤i≤d
j=1
(where γi,j is the j th component of γ i ) has a unique solution because the linear independence
of γ 1 , · · · , γ d implies that the matrix γi,j 1≤i,j≤d is invertible. So, for each α, there exists a
kα ∈ IRd such that kα · γ i = βα,γγ i for all 1 ≤ i ≤ d and hence
βα,γγ = kα · γ mod 2π
∀γγ ∈ Γ
Notice that, for each α, kα is not uniquely determined. Indeed
βα,γγ = kα · γ mod 2π
⇐⇒ (kα − k′α ) · γ ∈ 2πZZ
⇐⇒ kα − k′α ∈ Γ#
and
βα,γγ = k′α · γ mod 2π
∀γγ ∈ Γ
∀γγ ∈ Γ
Now relabel the eigenvalues and eigenvectors, replacing the index α by the corresponding value of k ∈ IRd /Γ# and another index n. The index n is needed because many kα ’s
with different values of α can be equal. Under the new labelling the eigenvalue/eigenvector
equations are
Hφn,k = en (k)φn,k
Tγ φn,k = eik·γγ φn,k
∀γγ ∈ Γ
(S.1)
The H–eigenvalue is denoted en (k) rather than en,k because, while k runs over the continuous
set IRd /Γ# , n will turn out to run over a countable set. Now fix any k and observe that
“Tγ φn,k = eik·γγ φn,k for all γ ∈ Γ” means that
φn,k (x + γ ) = eik·γγ φn,k (x)
for all x ∈ IRd and γ ∈ Γ. If the eik·γγ were not there, this would just say that φn,k is periodic
with respect to Γ. We can make a simple change of variables that eliminates the eik·γγ . Define
ψn,k (x) = e−ik·x φn,k (x)
c Joel Feldman.
2000. All rights reserved.
2
Then subbing φn,k (x) = eik·x ψn,k (x) into (S.1) gives
1
2m
2
i∇
∇ − k ψn,k + V ψn,k = en (k, V )ψn,k
(S.2)
ψn,k (x + γ ) = ψn,k (x)
Problem S.3 Prove that, for all ψ(x) in the obvious domains
i) (i∇
∇) eik·x ψn,k (x) = eik·x (i∇
∇ − k)ψn,k (x)
2 ik·x
ik·x
ii) (i∇
∇) e
ψn,k (x) = e
(i∇
∇ − k)2 ψn,k (x)
iii) V (x) eik·x ψn,k (x) = eik·x V (x)ψn,k (x)
iv) Tγ eik·x ψn,k (x) = eik·x eik·γγ Tγ ψn,k (x)
Denote by INk the set of values of n that appear in pairs α = (k, n) and define
Hk = span
φn,k n ∈ INk
Then, formally, and in particular ignoring that k runs over an uncountable set,
L2 (IRd ) = span
φn,k k ∈ IRd /Γ# , n ∈ INk = ⊕k∈IRd /Γ# Hk
Set
H̃k = span
ψn,k n ∈ INk
As multiplication by e−ik·x is a unitary operator, Hk is unitarily equivalent to H̃k and L2 (IRd )
is unitarily equivalent to ⊕k∈IRd /Γ# H̃k . The restriction of the Schrödinger operator H to H̃k
2
1
i∇
∇ − k + V applied to functions that are periodic with respect to Γ.
is 2m
So what have we gained? At least formally, we now know that to find the spectrum of
2
1
H = 2m
i∇
∇ +V (x), acting on L2 IRd , it suffices to find, for each k ∈ IRd /Γ# , the spectrum
2
1
i∇
∇ − k + V (x) acting on L2 IRd /Γ . We shall shortly prove that, unlike H,
of Hk = 2m
Hk has compact resolvent. So, unlike H (which we shall see has continuous spectrum), the
spectrum of Hk necessarily consists of a sequence of eigenvalues en (k) converging to ∞. We
shall also prove that the functions en (k) are continuous in k and periodic with respect to Γ#
and that the spectrum of H is precisely en (k) n ∈ IN, k ∈ IRd /Γ# .
Our next steps are to really prove that the spectrum of H is determined by the
spectra of the Hk ’s and then that the Hk ’s have compact resolvent.
c Joel Feldman.
2000. All rights reserved.
3
§II The Reduction from H to the Hk ’s
We now rigorously express H as a “sum” (technically a direct integral) of Hk ’s.
Because we are working in a rather concrete setting, we will never have to define what a
direct integral is. We shall make “L2 (IRd ) is unitarily equivalent to ⊕k∈IRd /Γ# H̃k ” rigorous
by constructing a unitary operator U from the space of L2 functions f (x), x ∈ IRd to the
space of L2 functions ψ(k, x), k ∈ IRd /Γ# , x ∈ IRd /Γ with the property that
(U HU ∗ ψ)(k, x) = Hk ψ(k, x)
Define
n
d
d ∞
S IR /Γ × IR /Γ = ψ ∈ C IR × IR d
#
d
ψ(k, x + γ ) = ψ(k, x) ∀γγ ∈ Γ
eib·x ψ(k + b, x) = ψ(k, x) ∀b ∈ Γ#
o
Define an inner product on S IRd /Γ# × IRd /Γ by
hψ, φiΓ =
1
|Γ# |
Z
dk
IRd /Γ#
Z
dx
ψ(k, x) φ(k, x)
IRd /Γ
With this inner product S IRd /Γ# × IRd /Γ is almost a Hilbert space. The only missing
axiom is completeness. Call the completion L2 IRd /Γ# × IRd /Γ .
Remark S.1 The condition ψ(k, x + γ ) = ψ(k, x) ∀γγ ∈ Γ just says that ψ is periodic
with respect to Γ in the argument x. The condition eib·x ψ(k + b, x) = ψ(k, x) ∀b ∈ Γ# , or
equivalently ei(k+b)·x ψ(k + b, x) = eik·x ψ(k, x) ∀b ∈ Γ# , says that eik·x ψ(k, x) is periodic
with respect to Γ# in the argument k. The extra factor eik·x means that ψ(k, x) itself need
not be periodic with respect to Γ# in the argument k. So ψ(k, x) need not be continuous on
the torus IRd /Γ# × IRd /Γ and my notation S IRd /Γ# × IRd /Γ is not very technically correct.
There is a fancy way of formulating the second condition as a continuity condition which leads
to the statement “ψ(k, x) is a smooth section of the line bundle . . . over IRd /Γ# × IRd /Γ”.
Remark S.2 On the other hand, if both ψ(k, x) and φ(k, x) are in S IRd /Γ# × IRd /Γ ,
then the integrand ψ(k, x) φ(k, x) is periodic with respect to Γ# in k and is periodic with
respect to Γ in x. Hence if D is any fundamental domain for Γ and D# is any fundamental
domain for Γ#
Z
Z
1
hψ, φiΓ = |Γ# | dk dx ψ(k, x) φ(k, x)
D#
c Joel Feldman.
2000. All rights reserved.
D
4
The value of the integral is independent of the choice of D and D# . Thus, you can always
realize L2 IRd /Γ# × IRd /Γ as the conventional L2 D# × D .
Also define
S IR
d
=
n
f ∈C
∞
Q
d
2n
IR sup (1 + x )
∂ ij
ij
j=1 ∂xj
d
x
f (x) < ∞
∀n, i1 , · · · id ∈ IN
o
This is called “Schwartz space”. A function f (x) is in Schwartz space if and only all of
its derivatives are continuous and decay, for large |x|, faster than one over any polynomial.
Think of S IRd as a subset of L2 (IRd . Set
(uψ)(x) =
(ũf )(k, x) =
1
|Γ# |
X
Z
dd k eik·x ψ(k, x)
IRd /Γ#
e−ik·(x+γγ ) f (x + γ )
γ ∈Γ
Proposition S.3
i) u : S IRd /Γ# × IRd /Γ → S IRd
ii) ũ : S IRd → S IRd /Γ# × IRd /Γ
iii) ũuψ = ψ for all ψ ∈ S IRd /Γ# × IRd /Γ
iv) uũf = f for all f ∈ S IRd
v) hũf, ũgiΓ = hf, gi for all f, g ∈ S IRd
vi) huψ, uφi = hψ, φiΓ for all ψ, φ ∈ S IRd /Γ# × IRd /Γ
vii) hf, uφi = hũf, φiΓ for all f ∈ S IRd , φ ∈ S IRd /Γ# × IRd /Γ
Proof:
i) This is Problem S.4 . It is the usual integration by parts game. Note that the integrand
eik·x ψ(k, x) is periodic with respect to Γ# in the integration variable k.
ii) Fix f ∈ S IRd and set
ψ(k, x) =
X
e−ik·(x+γγ ) f (x + γ )
γ ∈Γ
c Joel Feldman.
2000. All rights reserved.
5
const
1+|x|d+1
As f (x) and all of its derivatives are bounded by
X Q
d
∂ iℓ ∂ jℓ
i
j
∂xℓℓ ∂kℓℓ
γ ∈Γ ℓ=1
the series
e−ik·(x+γγ ) f (x + γ )
converges absolutely and uniformly in k and x (on any compact set) for all i1 , · · · , id , j1 , · · · jd .
Consequently ψ(k, x) exists and is C ∞ . We now verify the periodicity conditions. If γ ∈ Γ,
X
′
ψ(k, x + γ ) =
e−ik·(x+γγ +γγ ) f (x + γ + γ ′ )
γ ′ ∈Γ
X
=
′′
e−ik·(x+γγ ) f (x + γ ′′ )
where γ ′′ = γ + γ ′
γ ′′ ∈Γ
= ψ(k, x)
and, if b ∈ Γ# ,
ei(k+b)·x ψ(k + b, x) =
X
ei(k+b)·x e−i(k+b)·(x+γγ ) f (x + γ ) =
γ ∈Γ
=
X
e−i(k+b)·γγ f (x + γ )
γ ∈Γ
X
e−ik·γγ f (x + γ ) = eik·x
γ ∈Γ
X
e−ik·(x+γγ ) f (x + γ )
γ ∈Γ
ik·x
=e
ψ(k, x)
iii) Let
f (x) = (uψ)(x)
Z
1
|Γ# |
=
Ψ(k, x) = (ũf )(k, x) =
X
dd k eik·x ψ(k, x)
IRd /Γ#
e−ik·(x+γγ ) f (x + γ )
γ ∈Γ
Then
Ψ(k, x) =
X
−ik·(x+γ
γ)
e
1
|Γ# |
γ ∈Γ
Z
dd p eip·(x+γγ ) ψ(p, x + γ )
IRd /Γ#
so that, by the periodicity of ψ in γ ,
ik·x
e
Ψ(k, x) =
X
−ik·γ
γ
e
γ ∈Γ
1
|Γ# |
Z
dd p eip·(x+γγ ) ψ(p, x)
IRd /Γ#
Fix any x and recall that h(p) = eip·x ψ(p, x) is periodic in p with respect to Γ# . Hence by
Theorem L.10, (all labels “L.*” refer to the notes “Lattices and Periodic Functions”) with
Γ → Γ# , b → −γγ , f → h, x → p in the integral and x → k in the sum
Z
X
−iγ
γ ·k
1
dd p eiγγ ·p h(p)
h(k) = |Γ# |
e
γ ∈Γ
c Joel Feldman.
2000. All rights reserved.
IRd /Γ#
6
Subbing in h(p) = eip·x ψ(p, x)
ik·x
e
ψ(k, x) =
1
|Γ# |
X
−iγ
γ ·k
e
γ ∈Γ
Z
dd p eiγγ ·p eip·x ψ(p, x)
IRd /Γ#
so that eik·x Ψ(k, x) = eik·x ψ(k, x) and Ψ(k, x) = ψ(k, x), as desired.
iv) Let
ψ(k, x) = (ũf )(k, x) =
X
e−ik·(x+γγ ) f (x + γ )
γ ∈Γ
F (x) = (uψ)(x)
=
1
|Γ# |
Z
dd k eik·x ψ(k, x)
IRd /Γ#
Then
F (x) =
=
Z
1
|Γ# |
X
X
d
d k
IRd /Γ#
f (x + γ )
γ ∈Γ
−ik·γ
γ
e
f (x + γ ) =
γ ∈Γ
X
f (x + γ ) |Γ1# |
γ ∈Γ
Z
dd k e−ik·γγ
IRd /Γ#
1 if γ = 0
0 if γ 6= 0
= f (x)
v) Let
[γγ 1 , · · · γ d ] =
n P
d
j=1
o
tj γ j 0 ≤ tj ≤ 1 for all 1 ≤ j ≤ d
be the parallelepiped with the γ j ’s as edges.
Z
Z
1
dk
dx (ũf )(k, x) (ũg)(k, x)
hũf, ũgiΓ = |Γ# |
IRd /Γ
IRd /Γ#
Z
Z
1
= |Γ# |
dk
dx
(ũf )(k, x) (ũg)(k, x)
IRd /Γ#
[γ
γ 1 ,···,γ
γd]
Z
Z
hX
ihX
i
′
1
dk
dx
e−ik·(x+γγ ) f (x + γ )
= |Γ# |
e−ik·(x+γγ ) g(x + γ ′ )
IRd /Γ#
=
dx
Z
dx
Z
dx f (x) g(x)
[γ
γ 1 ,···,γ
γd ]
=
[γ
γ 1 ,···,γ
γd ]
=
[γ
γ 1 ,···,γ
γd]
Z
X
f (x + γ )g(x + γ ′ )
γ ,γ
γ ′ ∈Γ
X
γ ′ ∈Γ
γ ∈Γ
1
|Γ# |
Z
dk eik·(γγ −γγ
′
)
IRd /Γ#
f (x + γ ) g(x + γ )
γ ∈Γ
IRd
vi) Set f = uψ and g = uφ. Then, by part (iii), ũf = ψ and ũg = φ so that, by part (v),
huψ, uφi = hf, gi = hũf, ũgiΓ = hψ, φiΓ
c Joel Feldman.
2000. All rights reserved.
7
vii) Set g = uφ. Then, by part (iii), ũg = φ so that, by part (v),
hf, uφi = hf, gi = hũf, ũgiΓ = hũf, φiΓ
The mass m plays no role, so we set it to
1
2
from now on.
Proposition S.4 Let V be a C ∞ function that is periodic with respect to Γ and set
2
H = i∇
∇ + V (x)
2
Hk = i∇
∇x − k + V (x)
with domains S IRd and S IRd /Γ# × IRd /Γ , respectively. Then,
(ũHu ψ)(k, x) = (Hk ψ)(k, x)
for all ψ ∈ S IRd /Γ# × IRd /Γ
Observe that
i
h ik·x
i∇
∇x e
ψ(k, x) = −keik·x ψ(k, x) + eik·x i∇
∇x ψ (k, x) = eik·x [i∇
∇x − k]ψ (k, x)
R
As (uψ)(x) = |Γ1# | IRd /Γ# dd k eik·x ψ(k, x) , we have
Z
2
1
dd k eik·x ψ(k, x)
(Huψ)(x) = i∇
∇ + V (x) |Γ# |
d
#
IR /Γ
Z
d
ik·x
2
1
d ke
[i∇
∇x − k] ψ (k, x) + V (x)ψ(k, x)
= |Γ# |
Proof:
IRd /Γ#
= (uHk ψ)(x)
Now apply ũ to both sides and use Proposition S.3.iii.
Theorem S.5
i) The operators u and ũ have unique bounded extensions U : L2 IRd /Γ# ×IRd /Γ → L2 IRd
and Ũ : L2 IRd → L2 IRd /Γ# × IRd /Γ and
Ũ U = 1l
L2 IRd /Γ# ×IRd /Γ
U Ũ = 1l
L2 IRd
Ũ = U ∗
U = Ũ ∗
ii) The operators H (defined on S IRd ) and Hk (defined on S IRd /Γ# ×IRd /Γ ) have unique
self–adjoint extensions to L2 IRd and L2 IRd /Γ# × IRd /Γ . We also call the extensions H
and Hk . They obey
U ∗ HU = Hk
c Joel Feldman.
2000. All rights reserved.
8
Proof: i) ũ and u are bounded by Proposition S.3 parts (v) and (vi) respectively. As
S IRd /Γ# × IRd /Γ is dense in L2 IRd /Γ# × IRd /Γ and S IRd is dense in L2 IRd , ũ and u
have unique bounded extensions Ũ and U . The remaining claims now follow from Proposition
S.3 parts (iii), (iv), (vii) and (vii) respectively, by continuity.
2
ii) Step 1: i∇
∇ is essentially self–adjoint on the domain S IRd
The Fourier transform is a unitary map from L2 (IRd ) to L2 (IRd ), that maps S(IRd )
onto S(IRd ). Under this unitary map −∆, with domain S IRd , becomes the multiplication
operator Mx2 : S(IRd ) ⊂ L2 (IRd ) → L2 (IRd ) defined by Mx2 ϕ(x) = x2 ϕ(x). So it suffices to
prove that the operator Mx2 is essentially self–adjoint on S(IRd ). But if ϕ(x) ∈ S(IRd ), then
ϕ(x)
d
x2 ±i ∈ S(IR ).
d
2
Hence the range of Mx2 ± i contains all of S(IRd ) and consequently is dense
in L (IR ). Now just apply the Corollary of [Reed and Simon, volume I, Theorem VIII.3].
ii) Step 2: Prove Lemma S.6, below.
ii) Step 3: Finish off the proof.
2
In step 1, we saw that i∇
∇ is essentially self–adjoint on S IRd . The multiplication
operator V (x) is bounded and self–adjoint on L2 IRd . Consequently, by step 2, their sum,
H is essentially self–adjoint on S IRd and has a unique self–adjoint extension in L2 IRd .
The unitary operator U provides a unitary equivalence with
L2 IRd /Γ# × IRd /Γ ↔ L2 IRd
S IRd /Γ# × IRd /Γ ↔ S IRd
Hk ↾ S IRd /Γ# × IRd /Γ ↔ H ↾ S IRd
So Hk is essentially self–adjoint on S IRd /Γ# × IRd /Γ , has a unique self–adjoint extension
in L2 IRd /Γ# × IRd /Γ and the extensions obey U ∗ HU = Hk .
Lemma S.6 Let H be a Hilbert space and D a dense subspace of H. Let T : D → H be
essentially self-adjoint, with unique self–adjoint extension T̄ , and V : H → H be bounded
and self–adjoint. Then T + V : D → H is essentially self–adjoint and the unique self–adjoint
extension of T + V is T̄ + V .
Proof:
As V is bounded,
ϕ = lim ϕn , ψ = lim T ϕn
n→∞
n→∞
⇐⇒
ϕ = lim ϕn , ψ + V ϕ = lim (T + V )ϕn
n→∞
n→∞
for any sequence ϕ1 , ϕ2 , · · · ∈ D. The left hand side defines “ϕ ∈ DT , T ϕ = ψ” and the
right hand side defines “ϕ ∈ DT +V , T + V ϕ = ψ + V ϕ”, so T + V is closeable with closure
T +V =T +V.
c Joel Feldman.
2000. All rights reserved.
9
To prove that T + V is self–adjoint, it suffices to prove that, for any densely define
operator A and any bounded operator V , (A + V )∗ = A∗ + V ∗ . But, as V is bounded and
DA+V = DA ,
f ∈ DA∗ ⇐⇒
⇐⇒
sup
g∈DA ,kgk=1
sup
g∈DA ,kgk=1
hAg, f i < ∞
h(A + V )g, f i < ∞
⇐⇒ f ∈ D(A+V )∗
and, for all g ∈ DA = DA+V , f ∈ DA∗ = D(A+V )∗
hg, (A + V )∗ f i = h(A + V )g, f i = hAg, f i + hV g, f i = hg, A∗ f i + hg, V ∗ f i = hg, (A∗ + V ∗ )f i
§III Compactness of the Resolvent of Hk , for each fixed k
In this section we fix a lattice Γ in IRd , a vector k ∈ IRd and a smooth, real–valued,
function V (x) ∈ C ∞ (IRd /Γ) and study the operator
2
Hk = i∇
∇ − k + V (x)
acting on L2 IRd /Γ .
We denote by
(F f (b) = √1
|Γ|
(F −1 ϕ (x) = √1
|Γ|
Z
dd x e−ib·x f (x)
IRd /Γ
X
eib·x ϕ(b)
(S.3)
b∈Γ#
the Fourier transform and its inverse, normalized so that they are unitary maps from
L2 IRd /Γ to ℓ2 (Γ# ) and from ℓ2 (Γ# ) to L2 IRd /Γ respectively.
Lemma S.7
2
a) The operator i∇
∇ − k is self–adjoint on the domain
D=
F −1 ϕ ϕ(b), b2 ϕ(b) ∈ ℓ2 (Γ# )
and essentially self–adjoint on the domain
D0 =
c Joel Feldman.
F −1 ϕ ϕ(b) = 0 for all but finitely many b ∈ Γ#
2000. All rights reserved.
10
2
b) The spectrum of i∇
∇ − k − λ1l is
c) If 0 is not in
operator with norm
(b − k)2 − λ b ∈ Γ#
2
−1
(b − k)2 − λ b ∈ Γ# , i∇
∇ − k − λ1l
exists and is a compact
h
i−1
2
−1 ∇ − k − λ1l = min |(b − k)2 − λ|
i∇
#
b∈Γ
For d < 4, it is Hilbert-Schmidt.
a) Let
Proof:
ϕ ∈ ℓ2 (Γ# ) b2 ϕ(b) ∈ ℓ2 (Γ# )
D̃0 = ϕ ∈ ℓ2 (Γ# ) ϕ(b) = 0 for all but finitely many b ∈ Γ#
D̃ =
M = the operator of multiplication by (b − k)2 on D̃
m = the operator of multiplication by (b − k)2 on D̃0
If ϕ ∈ D̃0 then
ϕ(b)
(b−k)2 ±i
ϕ
is also in D̃0 so that ϕ = (m ± i) (b−k)
2 ±i is in the range of m ± i.
Thus the range of m ± i is all of D̃0 and hence is dense in ℓ2 (Γ# ). This proves that m is
essentially self–adjoint.
Recall that, since (α − β)2 ≥ 0, we have 2αβ ≤ α2 + β 2 for all real α and β. Hence
b2 = (b − k + k)2 = (b − k)2 + 2(b − k) · k + k2 ≤ (b − k)2 + 2kb − kk kkk + k2
≤ (b − k)2 + kb − kk2 + kkk2 + k2 = 2(b − k)2 + 2k2
ϕ(b)
ϕ
Consequently, if ϕ ∈ ℓ2 (Γ# ), then (b−k)
2 ±i ∈ D̃ so that ϕ = (M ± i) (b−k)2 ±i is in the range
of M ± i. Thus the range of M ± i is all of ℓ2 (Γ# ). This proves that M is self–adjoint and
hence is the unique self–adjoint extension of m.
2
The operator F i∇
∇ −k F −1 is the operator of multiplication by (b−k)2 on ℓ2 (Γ# ).
2
Hence i∇
∇ − k is self–adjoint on F −1 D̃ = D and essentially self–adjoint on F −1 D̃0 = D0 .
2
b) The operator i∇
∇ − k − λ1l is unitarily equivalent to the operator of multiplication by
(b − k)2 − λ on ℓ2 (Γ# ). The function (b − k)2 − λ takes the values (b − k)2 − λ b ∈ Γ# .
Each of these values is taken on a set of nonzero measure (with respect to the counting measure
on Γ# ). So the spectrum of (b − k)2 − λ contains
(b − k)2 − λ b ∈ Γ# .
In part c, below, we shall show that, if 0 is not in (b − k)2 − λ b ∈ Γ# , then
1
(b−k)2 −λ
is bounded uniformly in b. That is, 0 is not in the spectrum of multiplication by
c Joel Feldman.
2000. All rights reserved.
11
(b − k)2 − λ. This is all we need, because if µ is not in (b − k)2 − λ b ∈ Γ# , then 0
is not in (b − k)2 − λ′ b ∈ Γ# , with λ′ = λ + µ, so that 0 is not in the spectrum of
multiplication by (b − k)2 − λ′ and µ is not in the spectrum of multiplication by (b − k)2 − λ.
c) Fix any k and any λ ∈ C such that (b − k)2 − λ is nonzero for all b ∈ Γ# . Set
Cr = inf
(b − k)2 − λ b ∈ Γ# , |b| ≥ r
Since (b − k)2 ≥ 21 b2 − k2 , Cr ≥ 12 r 2 − k2 − λ so that limr→∞ Cr = ∞ and
h
i−1
o
n
1
2
1
1
=
inf
|(b
−
k)
−
λ|
,
<∞
=
max
max
sup (b−k)
2 −λ
(b−k)2 −λ Cr
#
b∈Γ#
|b|<r
b∈Γ
Let R and Rr be the operators on ℓ2 (Γ# ) of multiplication by
1
(b−k)2 −λ
1
(b−k)2 −λ
and
1 if |b| ≤ r
0 if |b| > r
i−1
h
respectively. Then R is a bounded operator, with norm minb∈Γ# |(b − k)2 − λ|
, Rr is a
finite rank operator and kR − Rr k = C1r converges to zero as r tends to infinity. This proves
2
that R is compact. As i∇
∇ − k − λ1l is unitarily equivalent to the multiplication operator
−1
(b − k)2 − λ, its inverse (i∇
∇ − k)2 − λ1l
is unitarily equivalent to R and is also compact,
with the same operator norm as R.
1
b ∈ Γ# and its set
Now restrict to d < 4. The spectrum of R is
2
(b−k) −λ
1
#
b
∈
Γ
.
To
prove
that
R is Hilbert-Schmidt, we must
of singular values is |(b−k)
2 −λ|
prove that
2
X 1
(b−k)2 −λ < ∞
b∈Γ#
Choose any b1 , · · · , bd such that
Γ# =
n1 b1 + · · · + nd bd n1 , · · · , nd ∈ ZZ
Let B be the d × d matrix whose (i, j) matrix element is bi · bj . For every nonzero x =
(x1 , · · · , xd ) ∈ Cd
x · Bx = |x1 b1 + · · · xd bd |2 > 0
since the bi , 1 ≤ i ≤ d are independent. Consequently, all of the eigenvalues of B are strictly
larger than zero. Let β be the smallest eigenvalue of B. Then
|x1 b1 + · · · xd bd |2 = x · Bx ≥ β|x|2
c Joel Feldman.
2000. All rights reserved.
12
2
Hence if b = n1 b1 + · · · + nd bd and n2 = (n1 , · · · , nd ) ≥
(b − k)2 − λ ≥ 1 b2 − k2 − |λ| ≥
2
β 2
n
2
4
β
k2 + |λ|
− k2 − |λ| ≥
so that
2
2
X X 1
1
(n1 b1 +···+nd bd −k)2 −λ (b−k)2 −λ =
β 2
n
2
− β4 n2 ≥ β4 n2
X
4 2
βn2 n∈ZZd
b∈Γ#
≤
X
n∈ZZd
n2 ≤ 4 (k2 +|λ|)
β
2
1
(n1 b1 +···+nd bd −k)2 −λ +
d
≤ # n ∈ ZZ n2 ≤
+
16
β2
2
4
β (k
X
+ |λ|)
n∈ZZd
n2 > 4 (k2 +|λ|)
β
max
n∈ZZd
n2 ≤ 4 (k2 +|λ|)
β
1
|n|4
2
1
(n1 b1 +···+nd bd −k)2 −λ n∈ZZd
n6=0
This is finite because d < 4 and we have assumed that (b − k)2 − λ does not vanish for any
b ∈ Γ# .
Lemma S.8 The following hold for all k ∈ IRd .
a) The operator Hk is self–adjoint on the domain
D=
F −1 ϕ ϕ(b), b2 ϕ(b) ∈ ℓ2 (Γ# )
and essentially self–adjoint on the domain
D0 =
F −1 ϕ ϕ(b) = 0 for all but finitely many b ∈ Γ#
−1
b) If λ is not in the spectrum of Hk , the resolvent Hk − λ1l
is compact. If d < 4 it is
Hilbert-Schmidt. If Im λ 6= 0 or λ < − supx |V (x)|, then λ is not in the spectrum of Hk .
c) Let R > 0. There is a constant C such that
1 Hk − Hk′ 1−∆ ≤ C|k − k′ |
for all k, k′ ∈ IRd with |k|, |k′ | ≤ R. The constant C depends on V and R, but is otherwise
independent of k and k′ .
d) Let R > 0 and λ < − supx |V (x)|. There is a constant C ′ such that
−1 −1 − Hk′ − λ1l ≤ C ′ |k − k′ |
Hk − λ1l
c Joel Feldman.
2000. All rights reserved.
13
for all k, k′ ∈ IRd with |k|, |k′ | ≤ R. The constant C ′ depends on V , λ and R, but is otherwise
independent of k and k′ .
e) Let c ∈ Γ# and define Uc to be the multiplication operator eic·x on L2 (IRd /Γ). Then Ub
is unitary and
Uc∗ Hk Uc = Hk+c
2
Proof: a) i∇
∇ − k is self–adjoint on D and essentially self–adjoint on D0 and V (x) is a
bounded operator on L2 IRd /Γ . Apply Lemma S.6.
−1
b) If λ is not in the spectrum of Hk , the resolvent Hk − λ1l
exists and is bounded. This is
just the definition of “spectrum”. As Hk is self–adjoint, its spectrum is a subset of IR. Now
2
−1
consider λ < − supx |V (x)|. As i∇
∇ − k − λ1l
is unitarily equivalent to multiplication
1
1
by (b−k)12 −λ1l ≤ |λ|
, it is a bounded operator with norm at most |λ|
. As λ < − supx |V (x)|,
2
−1
i∇
∇ − k − λ1l V has operator norm at most supx |V (x)|/|λ| < 1 and
1
1
1
1
1
1
=
=
≤
2
1
sup
|V
(x)|
x
Hk − λ1l
|λ|
|λ| − supx |V (x)|
1l + (i∇
1−
∇ − k − λ1l
∇−k)2 −λ1l V i∇
|λ|
Hence the spectrum of Hk is a subset of [− supx |V (x)|, ∞).
By the resolvent identity
−1 2
−1 −1
2
−1
Hk − λ1l
= i∇
∇ − k + 1l
− Hk − λ1l [V − (1 + λ)1l)] i∇
∇ − k + 1l
n
o
−1
2
−1
= 1l − Hk − λ1l [V − (1 + λ)1l)]
i∇
∇ − k + 1l
n
o
−1
The left factor 1l − Hk − λ1l [V − (1 + λ)1l)] is a bounded operator and, by Lemma S.7,
2
−1
the right factor i∇
∇ − k + 1l
is compact (Hilbert-Schmidt for d < 4), so the product is
compact (Hilbert-Schmidt for d < 4).
c) First observe that, by Problem S.6 below,
1
maps
1l−∆
d
2
all of L2 (IRd /Γ) into D so that
1
1
and Hk′ 1−∆
are both defined on all of L IR /Γ . Expanding gives
Hk 1−∆
Hk − Hk ′
1
1−∆
Hence F Hk − Hk′
=
2
2 1
1
′
2
′2
i∇
∇ − k − i∇
∇ − k′
1−∆ = − 2i(k − k ) · ∇ + k − k
1−∆
1
−1
1−∆ F
is the multiplication operator
2(k−k′ )·b+k2 −k′ 2
1+b2
The claim then follows from
2b+k+k′ 1+b2 ≤
c Joel Feldman.
2000. All rights reserved.
= (k − k′ ) ·
2|b|+2R
1+b2
≤
2b+k+k′
1+b2
1+b2 +2R
1+b2
≤ 1 + 2R
14
d) By the resolvent identity
1
1
1
1
Hk ′ − Hk
−
=
Hk − λ1l Hk′ − λ1l
Hk − λ1l
Hk′ − λ1l
1
1−∆
1
Hk ′ − Hk
=
Hk − λ1l
1 − ∆ (i∇
∇ − k′ )2 − λ1l 1l + V
1
1
(i∇
∇−k′ )2 −λ1l
By part c and the bound on the resolvent in part b,
1
|λ|
′ 1
1
1−∆
Hk −λ1l − Hk′ −λ1l ≤ |λ|−supx |V (x)| C|k − k | (i∇
∇−k′ )2 −λ1l |λ|−supx |V (x)|
As
1−∆
(i∇
∇−k′ )2 −λ1l
is unitarily equivalent to multiplication by
1+b2
,
(b−k′ )2 −λ
′
bounded uniformly on |k | < R.
1−∆
(i∇
∇−k′ )2 −λ1l
is
e) Since multiplication operators commute, Uc∗ V Uc = Uc∗ Uc V = V and the claim follows
immediately from Problem S.5, below.
Problem S.5 Let c ∈ Γ# and Uc be the multiplication operator eic·x on L2 (IRd /Γ). Let F
be the Fourier transform operator of (S.3).
a) Fill in the formulae
F Uc F −1 ϕ (b) =
F Uc∗ F −1 ϕ (b) =
ϕ(b
)
ϕ(b
)
b) Prove that Uc and Uc∗ both leave the domain D invariant.
c) Prove that
2
2
Uc∗ i∇
∇ − k Uc = i∇
∇−k−c
Problem S.6 Prove that
1
1l−∆
maps all of L2 (IRd /Γ) into D.
§IV The spectrum of H
We have just proven that the spectrum of the operator Hk (acting on L2 (IRd /Γ))
is contained in the half of the real line to the right of − supx |V (x)|. We have also just
−1
proven that the resolvent of Hk is compact. Hence the spectrum of Hk − λ1l
(for any
fixed suitable λ) is a sequence of eigenvalues converging to zero, so that the spectrum of Hk
consists of a sequence of eigenvalues converging to +∞. Denote the eigenvalues of Hk by
e1 (k) ≤ e2 (k) ≤ e3 (k) ≤ · · ·
c Joel Feldman.
2000. All rights reserved.
15
Proposition S.9
a) For each n, en (k) is continuous in k and periodic with respect to Γ# .
b) limn→∞ en (k) = ∞, with the limit uniform in k.
c) Denote by Vd the volume of a sphere of radius one in IRd . Let b1 , · · · , bd be any set of
Pd
− 1 ≤ tj < 1 for all 1 ≤ j ≤ d
generators for Γ# and B =
t
b
be the
j
j
j=1
2
2
parallelepiped, centered on the origin, with the bj ’s as edges. Denote by D the diameter of
B. For each k ∈ IRd and each R > 0
# n ∈ IN en (k) < R ≤
Vd
|Γ# |
# n ∈ IN en (k) < R ≥
Vd
|Γ# |
p
R + kV k + 21 D
d
=
Vd
Rd/2
|Γ# |
+O R
R − kV k − 21 D
d
=
Vd
Rd/2
|Γ# |
+O R
d−1
2
For each k ∈ IRd and each R > 14 D2 + kV k
p
d−1
2
This more detailed result concerning the rate at which en (k) tends to infinity with n is not
used in these notes and so may be safely skipped.
Proof:
2
b) Denote, in increasing order, the eigenvalues of i∇
∇−k
ê1 (k) ≤ ê2 (k) ≤ ê3 (k) ≤ · · ·
Each ên (k) is (b − k)2 , for some b ∈ Γ# . Furthermore, by Lemma S.7, the spectrum of
2
i∇
∇ − k is periodic in k, so that each ên (k) is periodic in k. We have already observed that
(b − k)2 ≥ 21 b2 − k2 , so that, as n tends to infinity, ên (k) tends to infinity, uniformly in k.
By the min-max principle
en (k) =
=
ên (k) =
inf
ϕ1 ,···,ϕn−1
ψ∈D, kψk=1
ψ⊥ϕ1 ,···,ϕn−1
hψ, Hk ψi
sup
inf
ϕ1 ,···,ϕn−1
ψ∈D, kψk=1
ψ⊥ϕ1 ,···,ϕn−1
ψ, (b − k)2 ψ + hψ, V ψi
sup
inf
ψ, (b − k)2 ψ
sup
ϕ1 ,···,ϕn−1
ψ∈D, kψk=1
ψ⊥ϕ1 ,···,ϕn−1
For any unit vector ψ, hψ, V ψi ≤ supx |V (x)|, so
en (k) − ên (k) ≤ sup |V (x)|
x
c Joel Feldman.
2000. All rights reserved.
16
and, as n tends to infinity, en (k) tends to infinity, uniformly in k.
a) Fix any λ < − supx |V (x)|. Denote, in increasing order, the eigenvalues of −[Hk − λ1l]−1
ẽ1 (k) ≤ ẽ2 (k) ≤ ẽ3 (k) ≤ · · ·
As
2 ϕ, [Hk − λ1l]ϕ = ϕ, i∇
∇ − k ϕ + ϕ, [V − λ1l]ϕ
Z
2
≥
[V (x) − λ]ϕ(x) dx
IRd /Γ
≥ |λ| − sup |V (x)| ϕ, ϕ
for all ϕ ∈ D, ẽn (k) < 0 and
x
en (k) = − ẽn1(k) + λ
for all n. Pick any R > 0. By Lemma S.8.d, for all unit vectors ϕ and all k, k′ with
|k|, |k′ | < R,
D E 1
1
ϕ, Hk −λ1l − Hk′ −λ1l ϕ ≤ C ′ |k − k′ |
1
1
Consequently, by the min-max principle, applied to A = − Hk −λ1
l and B = − Hk′ −λ1l
ẽn (k) − ẽn (k′ ) ≤ C ′ |k − k′ |
Hence, each ẽn (k), and consequently each en (k), is continuous.
2
c) By Lemma S.7, the spectrum of i∇
∇ − k is
(b − k)2 b ∈ Γ#
. Label these
2
eigenvalues, in order, f1 (k) ≤ f2 (k) ≤ f3 (k) ≤ · · ·. Observe that Hk and i∇
∇ − k both
have domain D and that, for every ϕ ∈ D,
D 2 E ∇ − k ϕ = hϕ, V ϕi ≤ kV k kϕk2
ϕ, Hk − i∇
Hence, by the min-max principle,
en (k) − fn (k) ≤ kV k
for all n and k so that, for all R > 0,
# n ∈ IN en (k) < R ≤ # n ∈ IN
# n ∈ IN fn (k) < R ≤ # n ∈ IN
fn (k) < R + kV k
en (k) < R + kV k
(S.4)
Let b + B be the half open parallelepiped, centered on b, with edges parallel to the bj ’s.
Then b + B b ∈ Γ# is a paving of IRd . This means that (b + B) ∩ (b′ + B) = ∅ unless
b = b′ and every point in IRd is in some b + B. So, for each r > 0
√ # n ∈ IN fn (k) < r = # b ∈ Γ# |b − k| < r
1
= |Γ# | Volume ∪b∈Sr b + B
c Joel Feldman.
2000. All rights reserved.
(S.5)
17
√ where Sr = b ∈ Γ# |b − k| < r .
Every point of b + B lies within distance of 12 D of b, so every point of ∪b∈Sr b + B
√
lies within a distance r + 12 D of k. On the other hand, if p ∈ IRd lies within a distance
√
r − 21 D of k, then p lies in precisely one b + B and that b obeys |p − b| ≤ 12 D and hence
√
√
|b − k| ≤ r − 12 D + 21 D ≤ r. Thus
Vd
√
r−
d
√
≤ Volume ∪b∈Sr b + B ≤ Vd r + 12 D
d
1
D
2
(S.6)
Subbing (S.6) in (S.5) gives
Vd
|Γ# |
√
r − 12 D
d
≤ # n ∈ IN fn (k) < r ≤
Vd
|Γ# |
√
r + 12 D
d
and subbing this into (S.4) gives the desired bounds.
Problem S.7 Let b1 , · · · , bd be any set of generators for Γ# and
B=
d
P
j=1
Prove that
tj b j −
1
2
≤ tj <
1
2
for all 1 ≤ j ≤ d
b + B b ∈ Γ# is a paving of IRd .
Theorem S.10 Let V be a C ∞ function of IRd that is periodic with respect to the lattice Γ
2
and H = i∇
∇ + V (x) the self–adjoint operator of Theorem S.5. The spectrum of H is
Proof:
en (k) k ∈ IRd /Γ# , n ∈ IN
Denote by ΣH the spectrum of H and by
S=
en (k) k ∈ IRd /Γ# , n ∈ IN
the set of all eigenvalues of all the Hk ’s.
Fix any p ∈ IRd and any n ∈ IN. We shall construct, for each ε > 0,
a vector ψε ∈ S IRd /Γ# × IRd /Γ obeying
Proof that S ⊂ ΣH :
c Joel Feldman.
ũHu − en (p)1l ψε ≤ εkψk
2000. All rights reserved.
18
−1
−1
This will prove that ũHu − en (p)1l
and hence H − en (p)1l
cannot be a bounded
−1
1
cannot be a
operator with norm at most ε , for any ε > 0; hence that H − en (p)1l
bounded operator and hence that en (p) ∈ ΣH .
By hypothesis, en (p) is an eigenvalue of Hp . So there is a nonzero vector ϕ̃(x) ∈ D
such that Hp − en (p) ϕ̃ = 0. As Hp is essentially self–adjoint on D0 , there is a sequence of
functions ϕm (x) ∈ D0 obeying
lim ϕm = ϕ̃
lim Hp ϕm = Hp ϕ̃
lim Hp − en (p)1l ϕm = 0
m→∞
=⇒
m→∞
lim kϕm k = kϕ̃k =
6 0
m→∞
m→∞
Hence there is a member of that sequence, call it ϕε (x), for which
Hp − en (p)1l ϕε ≤ ε kϕε k
2
Let f (k) be any nonnegative C ∞ function that is supported in
and whose square has integral one. Define, for each δ > 0,
fδ (k) =
k ∈ IRd |k| < 1
k ∈ IRd |k| < δ
1
f ( kδ )
δ d/2
Observe that fδ (k) is a nonnegative C ∞ function that is supported in
and whose square has integral one. Set
X
ψε (k, x) =
eic·x fδε (k − p + c)ϕε (x)
c∈Γ#
We shall choose δε later. The function ψε (k, x) is in S IRd /Γ# × IRd /Γ because
• the term fδε (k − p + c)ϕε (x) vanishes unless k is within a distance δε of p − c.
Hence ψε vanishes unless k ∈ Bδε (p − c), the ball of radius δε centered on p − c,
for some c ∈ Γ# . There is a nonzero lower bound on the distance between points of
Γ# . We will choose δε to be strictly smaller than half that lower bound. Then the
balls Bδε (p − c), c ∈ Γ# are disjoint. For k outside their union ψε (k, x) vanishes.
For k in Bδε (p − c0 ) the only term in the sum that does not vanish is that with
c = c0 . Consequently ψε (k, x) is C ∞ .
k
Bδε (p − c)
c Joel Feldman.
2000. All rights reserved.
19
• ψε (k, x) is periodic in x with respect to Γ because ϕε (x) is.
•
X
ψε (k + b, x) =
eic·x fδε (k + b − p + c)ϕε (x)
c∈Γ#
=
′
X
ei(c −b)·x fδε (k − p + c′ )ϕε (x)
c′ ∈Γ#
−ib·x
=e
ψε (k, x)
so ψε (k, x) has the required “twisted” periodicity in k.
The square of the norm
ũHu − en (p)1l ψε 2 =
1
|Γ# |
Z
Z
dk
2
dx (ũHu − en (p)1l)ψε (k, x)
IRd /Γ
IRd /Γ#
By Problem L.4 of the notes “Lattices and Periodic Functions”, we may choose
p+B =p+
d
P
j=1
tj b j −
1
2
≤ tj <
1
2
for all 1 ≤ j ≤ d
as the domain of integration in k. This domain contains the ball Bδε (p − c) with c = 0 and
does not intersect Bδε (p − c) for any c ∈ Γ# \ {0} (again assuming that δε has been chosen
p+B
sufficiently small). On p + B, ψε (k, x) = fδε (k − p)ϕε (x) and
(ũHu − en (p)1l)ψε (k, x) = fδε (k − p) Hk − en (p)1l)ϕε )(x)
so that
ũHu − en (p)1l ψε 2 =
=
1
|Γ# |
Z
1
|Γ# |
Z
The norm
c Joel Feldman.
dk
Z
2
dx fδε (k − p)2 Hk − en (p)1l)ϕε )(x)
IRd /Γ
2
dk fδε (k − p)2 Hk − en (p)1l)ϕε L2 (IRd /Γ,dx)
k Hk − en (p)1l)ϕε ≤ k Hp − en (p)1l)ϕε + k Hk − Hp )ϕε 1 k(1l − ∆)ϕε k
≤ 2ε kϕε + k Hk − Hp ) 1l−∆
≤ 2ε kϕε + C|k − p| k(1l − ∆)ϕε k
≤ 2ε kϕε + Cδε k(1l − ∆)ϕε k
2000. All rights reserved.
20
for k in the support of fδε (k − p). Now choose
kϕε k
ε
2C max{1,k(1l−∆)ϕε k}
δε =
With this choice of δε , k Hk − en (p)1l)ϕε L2 (IRd /Γ,dx) ≤ εkϕε L2 (IRd /Γ,dx) so that
Z
ũHu − en (p)1l ψε 2 ≤ ε#2
dk fδε (k − p)2 kϕε k2L2 (IRd /Γ,dx) = ε2 kψε k2
|Γ |
as desired.
Proof that ΣH ⊂ S: Fix any λ ∈
/ S. We must show that λ ∈
/ ΣH . As, for each fixed n,
en (k) is periodic and continuous in k,
inf en (k) − λ > 0
k
By Lemma S.9.b,
lim inf en (k) = ∞
n→∞ k
Hence
D = inf en (k) − λ > 0
k∈IRd
n∈IN
By the spectral theorem
ϕ 2 d
Hk − λ1l)ϕ 2 d
≥
D
L (IR /Γ,dx)
L (IR /Γ,dx)
for all ϕ in the domain, D, of Hk and in particular for all ϕ ∈ C ∞ IRd /Γ . Consequently, for
all ψ(k, x) ∈ S IRd /Γ# × IRd /Γ
Z
Z
2
1
(ũHu − λ1l)ψ (k, x)2
ũHu − λ1l ψ = #
dk
dx
|Γ |
IRd /Γ#
IRd /Γ
Z
Z
2
= |Γ1# |
dk
dx (Hk − λ1l)ψ (k, x)
IRd /Γ
IRd /Γ#
Z
2
dk (Hk − λ1l)ψ(k, · )L2 (IRd /Γ,dx)
= |Γ1# |
IRd /Γ#
Z
2
2
dk ψ(k, · )L2 (IRd /Γ,dx)
≥ |ΓD# |
IRd /Γ#
2
= D2 kψk
Recall that u is a unitary map from S IRd /Γ# × IRd /Γ onto S(IRd ). Hence
(H − λ1l)f ≥ Dkf k
(S.7)
for all f ∈ S(IRd ). By Theorem S.5, H is essentially self–adjoint on S(IRd ), so (S.7) applies
−1
1
for all f in the domain of H and H − λ1l
is a bounded operator with norm at most D
.
Hence λ is not in the spectrum of H.
c Joel Feldman.
2000. All rights reserved.
21
§V A Nontrivial Example – the Lamé Equation
Fix two real numbers β, γ > 0. The Weierstrass function with primitive periods γ
and iβ is the function ℘ : C → C defined by
℘(z) =
1
+
z2
1
1
− 2
2
(z − ω)
ω
X
ω∈γZZ⊕iβZZ
ω6=0
It is an elliptic function, which means that it is a meromorphic function that is doubly
periodic. It is analytic everywhere except for a double pole at each point of γZZ ⊕ iβZZ and
it has periods γ and iβ. The Weierstrass function is discussed in the notes “An Elliptic
Function – The Weierstrass Function”. The labels “W.*” refer to those notes. Two functions
closely related to ℘ are
Y
σ(z) = z
1−
ω∈γZZ⊕iβZZ
ω6=0
ζ(z) =
σ ′ (z)
σ(z)
=
1
z
+
z
ω
X
1 z2
z
e ω + 2 ω2
1
z−ω
+
1
ω
+
z
ω2
ω∈γZZ⊕iβZZ
ω6=0
As ζ ′ (z) = −℘(z), ζ is an antiderivative of −℘ and consequently is, except for some constants
of integration, periodic too. Similarly, σ is the exponential of an antiderivative of ζ and it is
not hard to determine how σ(z + γ) and σ(z + iβ) are related to σ(z).
Lemma W.4 There are constants η1 ∈ IR and η2 ∈ iIR satisfying
η1 iβ − η2 γ = 2πi
such that
ζ(z + γ) = ζ(z) + η1
η1 (z+ γ2
σ(z + γ) = −σ(z) e
ζ(z + iβ) = ζ(z) + η2
β
)
σ(z + iβ) = −σ(z) eη2 (z+i 2 )
Now set, for z ∈ C \ γZZ ⊕ iβZZ ,
ζ(z)x
ϕ(z, x) = e
λ(z) = −℘(z)
σ z − x − i β2
σ x + i β2
k(z) = −i ζ(z) − z ηγ1
ξ(z) = eγik(z) = eγζ(z)−zη1
c Joel Feldman.
2000. All rights reserved.
22
Lemma S.11
a)
ϕ(z, x + γ) = ξ(z) ϕ(z, x)
b)
β
d2
− dx
2 ϕ(z, x) + 2℘ x + i 2 ϕ(z, x) = λ(z)ϕ(z, x)
c)
ξ(z + γ) = ξ(z)
Proof:
ξ(z + iβ) = ξ(z)
a) By Problem W.3.d and Lemma W.4
ϕ(z, x + γ) = eζ(z)(x+γ)
σ z − x − γ − i β2
σ x + γ + i β2
σ − z + x + γ + i β2
= −e
σ x + γ + i β2
η (−z+x+i β + γ )
β
1
2
2
ζ(z)(x+γ) σ − z + x + i 2 e
= −e
β
γ
σ x + i β2 eη1 (x+i 2 + 2 )
β
ζ(z)(x+γ) −η1 z σ z − x − i 2
=e
e
σ x + i β2
ζ(z)(x+γ)
= eζ(z)γ−η1 z ϕ(z, x)
b) First observe that, since
β
d σ z −x−i2
dx σ x + i β2
′
σ z − x − i β2
σ ′ x + i β2 σ z − x − i β2
=−
+
σ z − x − i β2
σ x + i β2
σ x + i β2
h
iσ z − x − iβ = − ζ z − x − i β2 + ζ x + i β2
2
σ x + i β2
we have
d
ϕ(z, x)
dx
c Joel Feldman.
= ζ(z) − ζ z − x −
2000. All rights reserved.
i β2
−ζ x+
i β2
ϕ(z, x)
23
Differentiate again
d2
dx2 ϕ(z, x)
Lemma W.5 says that
ϕ(z, x)
= ζ z−x−
−ζ x+
2
+ ζ(z) − ζ z − x − i β2 − ζ x + i β2 ϕ(z, x)
β
β
= − ℘ z − x − i 2 − ℘ x + i 2 ϕ(z, x)
2
+ ζ(z) − ζ z − x − i β2 − ζ x + i β2 ϕ(z, x)
i β2
′
′
i β2
2
ζ(u + v) − ζ(u) − ζ(v) = ℘(u + v) + ℘(u) + ℘(v)
for all u, v ∈ C such that none of u, v, u + v are in γZZ ⊕ iβZZ (basically because, for each
fixed v ∈ C \ (γZZ ⊕ iβZZ), both the left and right hand sides are periodic and have double
poles, with the same singular part, at each u ∈ γZZ ⊕ iβZZ and each u ∈ −v + γZZ ⊕ iβZZ). By
this Lemma,
d2
dx2 ϕ(z, x)
β
β
= − ℘ z − x − i 2 − ℘ x + i 2 ϕ(z, x)
+ ℘(z) + ℘ z − x − i β2 + ℘ x + i β2 ϕ(z, x)
β
= ℘(z) + 2℘ x + i 2 ϕ(z, x)
c) By Lemma W.4,
ξ(z + γ) = eγζ(z+γ)−(z+γ)η1
ξ(z + iβ) = eγζ(z+iβ)−(z+iβ)η1
= eγζ(z)−zη1
= eγη2 −iβη1 eγζ(z)−zη1
= ξ(z)
= ξ(z)
Set Γ = γZZ and
V (x) = 2℘(x + i β2 )
d 2
+ V (x)
H = i dx
By Problem W.1, parts (b), (c) and (f), V ∈ C ∞ (IR/Γ) and is real valued. The Lamé equation
is
2
β
d
− dx
2 φ + 2℘(x + i 2 )φ = λφ
i.e. Hφ = λφ
(S.8)
A solution φ(k, x) of (S.8) that satisfies
φ(k, x + γ) = eiγk φ(k, x)
c Joel Feldman.
2000. All rights reserved.
(S.9)
24
is called a Bloch solution with energy λ and quasimomentum k.
Lemma S.11 says that, for each z ∈ C \ γZZ ⊕ iβZZ , ϕ(z, x) is a Bloch solution of
the Lamé equation with energy λ = λ(z) and quasimomentum k = k(z). I claim that the
energy λ and multiplier ξ = eγik are fully parameterized by
ξ(z) = eγζ(z)−zη1
λ(z) = −℘(z)
That is, the boundary value problem (S.8), (S.9) has a nontrivial solution if and only if
(λ, eiγk ) = (λ(z), ξ(z)), for some z ∈ C \ γZZ ⊕ iβZZ . The only if implication follows
from the observation, which is an immediate consequence of Lemma S.12 below, that unless
2z ∈ γZZ ⊕ iβZZ the functions ϕ(z, x) and ϕ(−z, x) are linearly independent solutions of (S.8)
for λ(z) = λ(−z). As a second order ordinary differential equation, (S.8) only has two linearly
independent solutions for each fixed value of λ. For z ∈ γZZ ⊕ iβZZ, λ(z) is not finite. For
2z ∈ γZZ ⊕ iβZZ with z ∈
/ γZZ ⊕ iβZZ, λ′ (z) = 0, by Corollary W.3, and the second linearly
∂
independent solution is ∂z
ϕ(z, x).
Lemma S.12
a) Let z1 , z2 ∈ C \ γZZ ⊕ iβZZ . If z1 − z2 ∈
/ γZZ ⊕ iβZZ, then ϕ(z1 , x) and ϕ(z2 , x) are linearly
independent (as functions of x).
b) If z ∈ C \ γZZ ⊕ iβZZ , then ϕ(z, x) and ∂∂z ϕ(z, x) are linearly independent (as functions
of x).
Proof: a) If ϕ(z1 , x) and ϕ(z2 , x) were linearly dependent, there would exist a, b ∈ C, not
both zero, such that aϕ(z1 , x) + bϕ(z2 , x) = 0 for all x ∈ IR. But
β
β
σ
z
−
x
−
i
σ
z
−
x
−
i
2
1
and
ϕ(z2 , x) = eζ(z2 )x
ϕ(z1 , x) = eζ(z1 )x
2
2
σ x + i β2
σ x + i β2
have analytic continuations to x ∈ C \ − i β2 + γZZ ⊕ iβZZ . These analytic continuations
must obey aϕ(z1 , x) + bϕ(z2 , x) = 0 for all x ∈ C \ − i β2 + γZZ ⊕ iβZZ . In particular, the
zero set of ϕ(z1 , x), which is z1 − i β2 + γZZ ⊕ iβZZ, must coincide with the zero set of ϕ(z2 , x),
which is z2 − i β2 + γZZ ⊕ iβZZ. This is the case if and only if z1 − z2 ∈ γZZ ⊕ iβZZ.
b) Fix any z ∈ C \ γZZ ⊕ iβZZ . As
β
′
∂ϕ
′
ζ(z)x σ z − x − i 2
= xζ (x)ϕ(z, x) + e
∂z
σ x + i β2
and σ has only simple zeroes, the zero set of
c Joel Feldman.
2000. All rights reserved.
∂ϕ
∂z
cannot coincide with that of ϕ.
25
Theorem S.13 Set
Λ1 = −℘( γ2 )
Λ2 = −℘( γ2 + i β2 )
Λ3 = −℘(i β2 )
Then Λ1 , Λ2 , Λ3 are real, Λ1 < Λ2 < Λ3 and the spectrum of H is [Λ1 , Λ2 ] ∪ [Λ3 , ∞).
Proof: If, for given values of λ and k, the boundary value problem (S.8), (S.9) has a
nontrivial solution and if k is real then λ is in the spectrum of H. We know that all such
λ’s are also real.
Imagine walking along the path in the z–plane that follows the four line segments
from 0 to γ2 to γ2 + i β2 to i β2 and back to 0. As ℘(z) = ℘(z̄), ℘(−z) = ℘(z) and ℘(z − γ) =
℘(z − iβ) = ℘(z) (this is part of Problem W.1.f), λ(z) = −℘(z) remains real throughout the
entire excursion. Near z = 0,
λ(z) = −℘(z) ≈ − z12
so λ starts out near −∞ at the beginning of the walk and moves continuously to +∞ at the
end of the walk. Furthermore, by Corollary W.3, which states, in part,
℘(z) = ℘(z ′ ) if and only if z − z ′ ∈ γZZ ⊕ iβZZ or z + z ′ ∈ γZZ ⊕ iβZZ.
λ never takes the same value twice on the walk, because no two distinct points z, z ′ on the
walk obey z + z ′ ∈ γZZ ⊕ iβZZ or z − z ′ ∈ γZZ ⊕ iβZZ.
• On the first quarter of the walk, from z = 0 to z = γ2 , λ(z) increases from −∞ to
Λ1 = −℘( γ2 ). But we cannot put these λ’s into the spectrum of H because, by Problem
W.5.e, k(z) is pure imaginary on this part of the walk. You might worry that k(z) might
happen to be exactly zero at some points of this first quarter of the walk. This could only
happen at isolated points, because k(z) is a nonconstant analytic function. If this were
to happen, the Lamé Schrödinger operator would have an isolated eigenvalue of finite
multiplicity. We have already seen that no periodic Schrödinger operator can have such
eigenvalues.
c Joel Feldman.
2000. All rights reserved.
26
• On the second quarter of the walk, from z = γ2 to z = γ2 + i β2 , λ(z) increases from Λ1 to
Λ2 = −℘( γ2 + i β2 ). By Problem W.5.d, k(z) is pure real on this part of the walk, so these
λ’s are in the spectrum of H.
• On the third quarter of the walk, from z = γ2 + i β2 to z = i β2 , λ(z) increases from Λ2 to
Λ3 = −℘(i β2 ). By Problem W.5.e, these λ’s do not go into the spectrum of H.
• On the last quarter of the walk, from z = i β2 back to zero, λ(z) increases from Λ3 to +∞.
By Problem W.5.d, these λ’s are in the spectrum of H.
For more information on the Lamè equation, see
Edward Lindsay Ince, Ordinary Differential Equations, Dover Publications,
1956, section 15.62.
Edmund Taylor Whittaker and George Neville Watson, A Course of Modern
Analysis, chapter XXIII.
c Joel Feldman.
2000. All rights reserved.
27